The problem is based upon, and hides a simple Lemma. For any secant of a parallelogram, the algebraic sum of the distances from one pair of opposite vertices of the parallelogram to that secant is equal to the algebraic sum of the distances from the other pair of opposite vertices of the parallelogram to that secant, both being equal to twice the distance from the centre of the parallelogram to that secant. Proof. The proof is immediate, by considering the similar ratios of the segments involved; no actual calculations are involved. $\quad \square$ Now, the whole set-up is made so that such parallelograms are created (the only reason points $M$ and $N$ are taken as feet of perpendiculars from $D$ is so that the use of orthocentres creates parallel lines). This creates parallelograms $NH_1MD$ and $CNH_2M$, so twice applying the Lemma yields \[\textrm{d}(M, AB) + \textrm{d}(N, AB) = \textrm{d}(D, AB) + \textrm{d}(H_1, AB),\] \[\textrm{d}(M, AB) + \textrm{d}(N, AB) = \textrm{d}(C, AB) - \textrm{d}(H_2, AB),\] whence $\textrm{d}(H_1, AB) + \textrm{d}(H_2, AB) = \textrm{d}(C, AB)$, leading to the equality of areas $\boxed{[AH_1BH_2]=[ABC]}$, independently of the position of $D$ on the side $AB$.

This problem is easily solved with vectors/complex numbers. Taking the midpoint of $DC$ as origin,we have $H_1=D+M+N$ and $H_2=C+M+N$ thus $(H_1-H_2)=(D-C)$. The area $AH_1BH_2$ is equal to half the magnitude of cross product $ \frac{1}{2}AB\times H_1H_2=\frac{1}{2} AB\times DC=[ABC]$.

my solution: $S_{AH_{1}BH_{2}}=S_{ANH_{1}MBH_{2}}-S_{ANH_{1}}-S_{BMH_{1}}= (S_{BMH_{2}}-S_{BMH_{1})+(S_{ANH_{2}}-S_{ANH{1}})+S_{MNC}+S_{MDN}=S_{AND}+S_{BMD}+S_{MNC}+S_{MDN}=S_{ABC}}$

horizon wrote: $S_{ANH_{1}MBH_{2}}-S_{ANH_{1}}-S_{BMH_{1}}=(S_{BMH_{2}}-S_{BMH_{1}})+(S_{ANH_{2}}-S_{ANH{1}})+S_{MNC}+S_{MDN}$ I don't understand this step .

AlexanderMusatov wrote: This problem is easily solved with vectors/complex numbers. Taking the midpoint of $DC$ as origin,we have $H_1=D+M+N$ and $H_2=C+M+N$ thus $(H_1-H_2)=(D-C)$. The area $AH_1BH_2$ is equal to half the magnitude of cross product $ \frac{1}{2}AB\times H_1H_2=\frac{1}{2} AB\times DC=[ABC]$. Why $S H1BH2A = \frac{1}{2} ABH1H2$

MinatoF wrote: horizon wrote: $S_{ANH_{1}MBH_{2}}-S_{ANH_{1}}-S_{BMH_{1}}=(S_{BMH_{2}}-S_{BMH_{1}})+(S_{ANH_{2}}-S_{ANH{1}})+S_{MNC}+S_{MDN}$ I don't understand this step . It's not hard to prove.$S_{MDN}=S_{MH_{1}N}$ because $DMH_{1}N$ is parallelogramm:$DM$ perpendicular BC, $NH_{1}$ perpendicular BC;$DN$ perpendicular AC, $MH_{1}$ perpendicular AC. Similarly, $S_{MCN}=S_{MNH_{2}}$. If you're looking to the picture, you can see all this step.

horizon wrote: my solution: $(S_{BMH_{2}}-S_{BMH_{1})+(S_{ANH_{2}}-S_{ANH{1}})+S_{MNC}+S_{MDN}=S_{AND}+S_{BMD}+S_{MNC}+S_{MDN}}$ I don't understand.

It is easy to prove NCMH2 is a parallelogram and therefore triangles NMH2 and MNC are congruent. Consequently, ND = MH1 (distances from respective vertices to the orthocenters). We also know angle (H2MH1) is right (it participates in a rectangle which is easy to see). Thus, triangles CND and H2MH1 are also congruent from which follows the identity CD = H2H1 and the fact that CD is parallel to H2H1. Now, we will express the area of AH1BH2 using its diagonals AB and H1H2. The area is SAH1BH2 = AB.H2H1. sin(a)/2 = AB.CD.sin(a)/2, where a = angle (BDC) which is also equal to the angle formed by H2H1 and AB (H2H1 parallel to CD). Let's work out now the area of ABC using AB, CD and angle (a). SABC = SADC + SBDC = AD.DC.sin(180-a)/2 + BD.CD.sin(a)/2 = AB.CD.sin(a)/2 which is the same => SABC = SAH1BH2 Q.E.D.

Reflect wrt midpt of $MN$ to get $A',B'$ from $A,B$ resp. Then the problem is to prove that area of $A'CB'D$ is constant. To see this, let $E$ be on $AB$ such that $CE\parallel A'B\parallel B'A$. Now slide vertices of $A'CB'D$ as follows, $D$ to $E$, $B'$ to $A$, $A'$ to $B$. Each slide preservs its area and brings it to $ABC$ hence done.

yashanglar manam shunday ishladim

As $MH_2 \perp ND$, $ND \perp CN$ and $NH_2 \perp DM$, $DM \perp CM$, we get that $CMH_2N$ is a parallelogram and in particular $CH_2$ and $MN$ bisect each other. Analogously, $MH_1 \perp CN$, $CN\perp ND$ and $NH_1 \perp CM$, $CM \perp DM$ so $DMH_1N$ is a parallelogram and $DH_1$ and $MN$ bisect each other. Hence $CH_1$ and $DH_2$ also bisect each other, so $CDH_2H_1$ is a parallelogram. Therefore $S_{AH_1BH_2} = \frac{1}{2} AB \cdot H_1H_2 \cdot \sin \angle(AB, H_1H_2) = \frac{1}{2} AB \cdot CD \cdot \sin \angle ADC = \frac{1}{2} AB \cdot \mbox{dist}(C,AB) = S_{ABC}$ which clearly does not depend on $D$.